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### #1 bonanova

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Posted 28 March 2008 - 07:40 AM

Commemorating the post that matches my member number,
let's try a bit of useless [this side of Vegas at least] calculations.

In a standard deck of cards there are 52 cards: 1-10, J Q K of four suits S H D C.
In draw poker, 5 cards are dealt to each player. Hands are valued in increasing degree as follows.
Edit: Examples of each type of hand shown in red.

0 - High card: Highest ranking card. This hand is none of the following types. - 3D 6H 8C JH KD
1 - Pair: 2 cards of same rank; 3 cards of different ranks - 4C 4D x x x
2 - Two Pair: 2 cards of one rank, 2 cards of another rank, 1 card of different rank. - 6H 6S 9D 9H x
3 - Three of a kind: 3 cards of same rank; 2 cards of different ranks. - KC KD KS x x
4 - Straight: 5 cards in rank sequence, but not all of same suit. - 3D 4D 5C 6H 7S
5 - Flush: 5 cards all of the same suit, but not all in sequence. - 3H 6H 7H 9H JH
6 - Full House: 3 cards of one rank; 2 cards of another rank. - AH AC AS 8S 8H
7 - Four of a kind: 4 cards of same rank. 7S 7H 7D 7C x
8 - Straight flush: 5 cards in sequence and all of same suit. - 4H 5H 6H 7H 8H.

Prove that the valuation of 0-8 is reasonable by computing the probabilities of being dealt each type of hand.
Express the probabilities as fractions where the denominator is the number of possible unique 5-card hands dealt from a single deck.
Easy check - they add up to unity.

The prize for being first with all correct answers is 10 "attaboy"s.
The current value of an "attaboy" is that you may combine 100 of them with \$0.85 for a cup of coffee at Dunkin Donuts.
• 0

Vidi vici veni.

### #2 Noct

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Posted 28 March 2008 - 08:22 AM

Commemorating the post that matches my member number,
let's try a bit of useless [this side of Vegas at least] calculations.

In a standard deck of cards there are 52 cards: 1-10, J Q K of four suits S H D C.
In draw poker, 5 cards are dealt to each player. Hands are valued in increasing degree as follows.

0 - High card: Highest ranking card. This hand is none of the following types. - e.g. 3D 6H 8C JH KD
1 - Pair: 2 cards of same rank; 3 cards of different ranks - 4C 4D x x x
2 - Two Pair: 2 cards of one rank, 2 cards of another rank, 1 card of different rank. - 6H 6S 9D 9H x
3 - Three of a kind: 3 cards of same rank; 2 cards of different ranks. - KC KD KS x x
4 - Straight: 5 cards in rank sequence, but not all of same suit. - 3D 4D 5C 6H 7S
5 - Flush: 5 cards all of the same suit, but not all in sequence. - 3H 6H 7H 9H JH
6 - Full House: 3 cards of one rank; 2 cards of another rank. - AH AC AS 8S 8H
7 - Four of a kind: 4 cards of same rank. 7S 7H 7D 7C x
8 - Straight flush: 5 cards in sequence and all of same suit. - 4H 5H 6H 7H 8H.

Prove that the valuation of 0-8 is reasonable by computing the probabilities of being dealt each type of hand.
Express the probabilities as fractions where the denominator is the number of possible unique 5-card hands dealt from a single deck.
Easy check - they add up to unity.

The prize for being first with all correct answers is 10 "attaboy"s.
The current value of an "attaboy" is that you may combine 100 of them with \$0.85 for a cup of coffee at Dunkin Donuts.

What are you asking us? the probability of getting your proposed hands?

How I'm understanding it, the answer to what you're asking for 0 would be:
5/52 * 4/51 * 3/50 * 2/49 * 1/48

Those are the chances of getting 3D 6H 8C JH KD

Edited by bonanova, 28 March 2008 - 10:05 AM.

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### #3 Seventh Sage

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Posted 28 March 2008 - 12:38 PM

Commemorating the post that matches my member number,
let's try a bit of useless [this side of Vegas at least] calculations.

In a standard deck of cards there are 52 cards: 1-10, J Q K of four suits S H D C.
In draw poker, 5 cards are dealt to each player. Hands are valued in increasing degree as follows.
Edit: Examples of each type of hand shown in red.

0 - High card: Highest ranking card. This hand is none of the following types. - 3D 6H 8C JH KD
1 - Pair: 2 cards of same rank; 3 cards of different ranks - 4C 4D x x x
2 - Two Pair: 2 cards of one rank, 2 cards of another rank, 1 card of different rank. - 6H 6S 9D 9H x
3 - Three of a kind: 3 cards of same rank; 2 cards of different ranks. - KC KD KS x x
4 - Straight: 5 cards in rank sequence, but not all of same suit. - 3D 4D 5C 6H 7S
5 - Flush: 5 cards all of the same suit, but not all in sequence. - 3H 6H 7H 9H JH
6 - Full House: 3 cards of one rank; 2 cards of another rank. - AH AC AS 8S 8H
7 - Four of a kind: 4 cards of same rank. 7S 7H 7D 7C x
8 - Straight flush: 5 cards in sequence and all of same suit. - 4H 5H 6H 7H 8H.

Prove that the valuation of 0-8 is reasonable by computing the probabilities of being dealt each type of hand.
Express the probabilities as fractions where the denominator is the number of possible unique 5-card hands dealt from a single deck.
Easy check - they add up to unity.

The prize for being first with all correct answers is 10 "attaboy"s.
The current value of an "attaboy" is that you may combine 100 of them with \$0.85 for a cup of coffee at Dunkin Donuts.

I see a trick to this that might catch a few people, but computing probabilities like this might take me a while.
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### #4 Dackombe

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Posted 28 March 2008 - 02:09 PM

Long term reader, first time poster...

Spoiler for Denominator

Some hands are easier than others, might take me a while to calculate them all, I think the key challenge here is
Spoiler for challenge

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### #5 andreay

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Posted 28 March 2008 - 02:50 PM

Long term reader, first time poster...

Spoiler for Denominator

Some hands are easier than others, might take me a while to calculate them all, I think the key challenge here is
Spoiler for challenge

but has that been wroked out not including the order, e.g. you could pick 1h, 5s, 3c, jd, 9d but you could also pick these 5 cards in the order, 5s, jd, 1h, 9d, 3c?
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### #6 bonanova

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Posted 28 March 2008 - 03:37 PM

Dackombe has the denominator. Welcome to the Den!

The numerators are the challenge, for the reasons he mentioned.

Hint: you can discuss the probabilities one at a time, as if it were
eight puzzles. You might be surprised which one is the most
complicated to describe and calculate. Have fun with this.

To me, the interesting part is that it's fairly easy to look at five
cards and not confuse 4 of a kind with 2 pairs for example.
But if I'm a little bit sloppy with my math, I might count it that way.
Or even worse, count 4 of a kind simply as a pair.

What's needed is to describe what all 5 cards must or must not be.
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Vidi vici veni.

### #7 GIJeff

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Posted 28 March 2008 - 03:37 PM

but has that been wroked out not including the order, e.g. you could pick 1h, 5s, 3c, jd, 9d but you could also pick these 5 cards in the order, 5s, jd, 1h, 9d, 3c?

I would think that the probability of getting the hand would be the same, regardless of order
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### #8 Dackombe

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Posted 28 March 2008 - 03:59 PM

I think the hands with the fewest possibilites are the easiest to work out so I'll start with the straight flush and simply count them...
Spoiler for Straight Flush

Also I'll just count the Four of a kinds...
Spoiler for Four of a kind

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### #9 Dackombe

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Posted 28 March 2008 - 06:26 PM

Now it gets a little more complicated...
Spoiler for Full House

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### #10 andreay

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Posted 28 March 2008 - 08:14 PM

sorry i've been working on this all day:P cba to post working although if required ill try and remember them